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Problem 108: If the probability density of a random variable is given by; find the value k and the probability that the random variable takes on a value

Problem 108: If the probability density of a random variable is given by
$f(x)= \begin{array}{cc} \Bigg\{ & \begin{array}{cc} kx^3 & 0\le x\le1 \\ 0 & elsewhere \end{array} \end{array} $
find the value k and the probability that the random variable takes on a value
a. betweeen $\frac{1}{4}$ and $\frac{3}{4}$; b. greater than $\frac{2}{3}.$
Solution:
To find k, we must integrate $f(x)$ from $x=0$ to $x=1$ and set it equal to 1. Thus
$\int _{0}^{1}f(x)dx=1\Rightarrow \int _{0}^{1}k{x}^{3}dx=1\Rightarrow k\frac{{x}^{4}}{4}\Big|_0^1 =1 \Rightarrow k/4=1 \Rightarrow k =4$
a. betweeen $\frac{1}{4}$ and $\frac{3}{4}$;
$P(0.25\le X \le 0.75) = \int _{0.25}^{0.75}4{x}^{3}dx = x^4\Big|_{0.25}^{0.75} = 0.75^4-0.25^4 = 0.3125$

b. greater than $\frac{2}{3}.$ $P(X > \frac{2}{3}) = \int _{\frac{2}{3}}^{\1}4{x}^{3}dx = x^4\Big|_{\frac{2}{3}}^{1} =1^4 - \frac{2}{3}^4 = 0.802469$

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